HW4 Solution

Zachary Lovering
Nov. 7, 2008
MAE 2
HW#4
Homework 4 Solution
Problem 6.1
Description:
Consider an airplane patterned after the twin-engine Beechcraft Queen Air executive transport. The airplane
weight is 38220 , wing area is 27.3 , aspect ratio is 7.5, Oswald efficiency factor is 0.9, and zero-lift drag
coefficient is , 0.03. Calculate the thrust required to fly at a velocity of 350 ⁄ at (a) standard sea level
and (b) an altitude of 4.5 .
Given:
38220
27.3 7.5
0.9
, 0.03
350 ⁄
Find:
a) at standard sea level
b) an altitude of 4.5
Assumptions:
1. Steady level flight
2. Incompressible
Equations:
,
(6.19)
Analysis:
To find the required thrust, the definition of dynamic pressure is substituted into equation 6.19,
,
2
1
2
,
Numerical Substitution:
For part (a), it is first necessary to find the air density at sea level. From Appendix A,

1.2250
Additionally, must be represented in consistent units,
350
1000
1
1 3600
97.2
Using the equation derived in the analysis, the required thrust is given by,
Note that,
1
1.2250
2 97.2
·
4739.4 436.04
·
5175.4
27.3 0.03
1 1
238220
1.2250
97.2
27.30.97.5 ·
For part (b), it is first necessary to find the air density at 4.5 . From Appendix A,
0.77704
Using the equation derived in the analysis, the required thrust is given by,
1
0.77704
2 97.2
·
3006.3 687.42
·
3693.7
27.3 0.03
238220
0.77704
97.2
27.30.97.5

Problem 6.2
Description:
An airplane weighing 5000 is flying at standard sea level with a velocity of 200 ⁄ .
At this velocity the
⁄ ratio is a maximum. The wing area and aspect ratio are 200 and 8.5, respectively. The Oswald
efficiency factor is 0.93. Calculate the total drag on the airplane.
Given:
5000 200 ⁄
200 8.5
0.93
Find:
Total drag,
Assumptions:
1. Steady level flight
2. Incompressible
3. Sea level standard air
4. Flying at maximum ⁄
Equations:
,
(6.1c)
(5.17)
(5.20)
Analysis:
For a complete airplane, the drag coefficient, , is given by
As noted on pg. 461, when flying at maximum ⁄ ,
Hence,
,
, ,
2
Substituting this equation into the equation for drag for the entire plane yields,
2

Since the aircraft is in steady level flight . The equation for the lift coefficient, , is given by,
Substituting this equation into the drag equation,
2
2
Substituting in the definition of dynamic pressure, , yields
2
1
2
4
Numerical Substitution:
It is first necessary to find the air density at sea level. From Appendix B,
0.0023769
Additionally, must be represented in consistent units,
200
293.3
1
5280
1 3600
Using the equation derived in the analysis, the total drag is given by,
Note that,
45000
0.938.5 0.0023769
293.3
200
98.47
·
98.5 1 1
·

Problem 6.4
Description:
Consider an airplane patterned after the Beechcraft Bonanza V-tailed, single-engine light private airplane. The
characteristics of the airplane are as follows: 6.2, 181 , 0.91, 3000 , , 0.027.
The airplane is powered by a single piston engine of 345 maximum at sea level. Assume the power of the
engine is proportional to the free-stream density. The two-blade propeller has an efficiency of 0.83.
a) Calculate the power required at sea level
b) Calculate the maximum velocity at sea level
c) Calculate the power required at 12000 altitude
d) Calculate the maximum velocity at 12000 altitude
Given:
6.2
181
0.91
3000
, 0.027
345 hp
0.83
Find:
a) at sea level
b) at sea level
c) , at 12000
d) at 12000
Assumptions:
1. Steady level flight
2. Standard air
3. Incompressible
Equations:
(6.24)
,
(Required thrust from Prob. 6.1)
(6.31)
, ,
/
/
Analysis:
(6.39)
(6.38)
On pg. 419 it is stated that, “the condition that holds at minimum power required is ,
,.” Using this
relation, the minimum power flight velocity may be calculated as follows:

, 1
3 ,
,
3
3,
2
3 ,
4
3
,
4
3
,
Note that this problem did not specifically ask for the minimum power required at sea level. Hence, other
selected values of velocity can be selected and will yield different power requirements. These will be plotted in
the Numerical Substitution section. The required power is given by,
Substituting the equation derived in Problem 6.1 for the required thrust, , into this equation for required
power yields,
2
/
1
2
,
The maximum velocity, , will occur when the aircraft it at full power. The maximum power available is given
by,
This equation can be substituted into the equation for the required power, , and then solved for . However,
the resulting equation is a fourth-order polynomial and, although closed-form solutions to this problem do exist,
the simplest way to solve this problem is by graphical means. This will be done in the Numerical Substitution
section. Finally, to find the required power and corresponding velocity at a different altitude, the following
equations are used:
, ,
/
/
where , is the required power at sea level, which we found in part (a), is the sea level air density, and is
the maximum velocity at sea level. To solve for the maximum velocity at a different altitude, the available power
must first be adjusted. It is given that the power available is proportional to the air density and so the following
equation must hold:

, ,
As stated in part (b), the maximum velocity corresponding to this maximum power will be determined
graphically.
Numerical Substitution:
To find the power required at sea level, it is required to know the air density at sea level. From Appendix B, the
sea level standard air density is
0.0023769
Using the required power equation and the maximum and minimum power equations, the following graph was
produced:
Power Required [HP]
400
350
300
250
200
150
100
50
0
Sea Level Power Required vs. Velocity
0 50 100 150 200 250 300 350
Velocity [ft/s]
Power Required Maximum Velocity Minimum Power
For part (a), acceptable answers for required power lie on the “Power Required” curve. To find the minimum
power, the velocity at minimum power equation is used.
43000
30.91 6.20.027 0.0023769
181
220
192.5
165
137.5
110
82.5
55
27.5
0
Power Required [1000 lbf * ft/s]

36000000
0.26573
36000000
0.26573
36000000
0.26573
107.9
As can be seen, this result for agrees with the “Sea Level Power Required vs. Velocity” graph. Using this
velocity, the minimum power required is given by,
1
0.0023769 107.9
2
1810.027 23000
0.0023769
107.9
1810.91 6.2
·
7296
21877 ·
7296 ·
21877 ·
·
29173 ·
The required power can also be represented in hp as follows:
29173 ·
53.04 hp
1 hp
550 ·
As can be seen, this result for agrees with the “Sea Level Power Required vs. Velocity” graph. This concludes
that answer to part (a). In part (b), it is required to find the maximum velocity. To do so, it is first necessary to
find the maximum power available.
0.83345 hp
286.35 hp
Using the graph, this maximum power at sea level corresponds to

295
This value is represented by the “maximum velocity” line on the graph. For part (c), a new plot can be created to
adjust for the density difference at 12000 altitude. The air density at 12000 can be found in Appendix B.
0.0016480
Answers for required power lie on the “Power Required” curve in the graph below.
Power Required [HP]
300
250
200
150
100
50
0
12,000 ft Altitude Power Required vs. Velocity
0 50 100 150 200 250 300 350
Velocity [ft/s]
Maximum Power Maximum Velocity Minimum Power
The minimum power required can be solved as in part (a) or can be solved graphically. Additionally, the density
change can be accounted for as follows:
0.0023769
0.0016480
107.9
1.201
129.6
107.9
Similarly, the corresponding minimum power at altitude is,
165
137.5
110
82.5
55
27.5
0
Power Required [1000 lbf * ft/s]

, 29173 ·
35035 ·
0.0023769
0.0016480
63.70 hp
Since the engine output power is proportional to the air density, the power available must also be adjusted.
0.0016480
, 286.35 hp
0.0023769
, 286.35 hp0.6933
, 198.54 hp
Using the graph, this maximum power at 12000 corresponds to
289

Problem 6.25
Description:
The Predator UAV has the following characteristics: 14.85 , 11.45 , 1020 ,
295 . The power plant is a Rotax four-cylinder, four-stroke engine of 85 hp driving a two-blade, variable-
pitch pusher propeller. Assume 0.7, , 0.03, 0.9, and the 0.2 · ⁄ hp.
Calculate the
maximum velocity of the Predator at sea level.
Given:
14.85
11.45
1020
295 85 hp
0.7
, 0.03
0.9
0.2 · ⁄ hp
Find:
Assumptions:
1. Sea level standard air
2. Incompressible
3. Steady level flight
Equations:
(6.31)
,
(6.29)
Analysis:
The maximum velocity occurs when using maximum available power. To find the maximum available power, the
following equation is used:
This equation is substituted into the required power equation.
2
1
2
,
Finally, the definition of the aspect ratio is substituted in as follows:
1
2
,
2
⁄

1
2
,
2
This equation must be solved for . However, as noted in Problem 6.4, this equation is a fourth-order
polynomial and, although closed-form solutions to this problem do exist, the simplest way to solve this problem
is by graphical means or by iteration. The method of solving graphically was illustrated in Problem 6.4 and so the
method of iteration will be demonstrated in the Numerical Substitution section of this problem.
Numerical Substitution:
To begin, it is first required to find the density at sea level. From Appendix A, the sea level standard air density is
Next, will be converted into consistent SI units.
1.225
9.8
1020
1 9996
Similarly, the power, , must also be converted into consistent SI units.
Using the equation derived in the analysis,
746 W
85 hp
1 hp
63410
0.963410 1
1.225
2 11.450.03
57069 0.2104
33.65
57069 0.2104
33.65
0.2104
33.65 ·
·
57069 0
29996
1.225
14.85 0.7
First, note that the units are consistent. In other words, for the first term on the left hand side,
For the second term on the left hand side,
·
·
·
·
·

To solve this equation iteratively, first guess a value for , then calculate the left hand side. If it is equal to
zero then you have the correct value. Otherwise, guess a new and calculate the left hand side again. This
process is most easily done using a program like Excel since it is easy to store and solve an equation.
Alternatively, many calculators or programs such as Matlab have built in solvers. The solution by any of these
means will yield,
64.73